The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation- t - -x - 3 where x is in meters and t is in seconds- The work done by the force in the first 6 seconds is

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Standard XI

Physics

Work Done When Force Is Varying

The displacem...

Question

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation, t = $\sqrt{x}$ + 3 where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

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Solution

The correct option is C
0 J

$x = (t - 3)^{2} \Rightarrow v = \frac{d x}{d t} = 2 (t - 3)$

at t = 0; $v_{1} = - 6 m / s a n d a t t = 6 s e c, v_{2} = 6 m / s$

So, change in kinetic energy

$= W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} = 0$

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The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation, t = √x + 3 where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

The correct option is C 0 J x=(t−3)2⇒v=dxdt=2(t−3) at t = 0; v1=−6m/s and at t=6sec,v2=6m/s So, change in kinetic energy =W=12mv22−12mv21=0

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation, t = $\sqrt{x}$ + 3 where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

The correct option is C
0 J

$x = (t - 3)^{2} \Rightarrow v = \frac{d x}{d t} = 2 (t - 3)$

at t = 0; $v_{1} = - 6 m / s a n d a t t = 6 s e c, v_{2} = 6 m / s$

So, change in kinetic energy

$= W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} = 0$